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Fermat's Poser

Find the point whose sum of distances from the vertices (corners) of a given triangle is a minimum.

Cut-up Square

Age 14 to 16 Short Challenge Level:

First, we label the corners of the square, the midpoint, and the point at which the two lines intersect $A$, $B$, $C$, $D$, $E$ and$F$ respectively. We also label the four triangles created within the square as 1, 2, 3 and 4.

Call the area of triangle a.

Angle $AFD$ is equal to angle $EFC$ as they are vertically opposite, and angle $FAD$ is equal to angle $FCD$ as they are alternate. So, triangles 1 and 3 are similar. As $E$ marks the midpoint of a side of the square, length $CE$ is half the value of length $AD$.

This means that the area of triangle 3 is equal to 4a, as its base and height are double that of triangle 1. 

It also means that length $AF$ is equal to twice the value of length $FC$. If we let $AF$ be the base of triangle 3 and $FC$ be the base of triangle 2, then these triangles have the same height (equal to the perpendicular distance from $D$ to the line $AF$) so the area of triangle 2 is equal to half that of triangle 3 - that is, 2a.

As line $AC$ is a diagonal, the combined areas of triangles 2 and 3 is equal to half the area of the square. 4a + 2a = 6a, so the square has a total area of 12a. The combined area of triangle 1 and triangle 4 is then also 6a. As we have defined the area of triangle 1 as a, we then know that triangle 4 has an area of 5a.

As the ratio of $P:Q$ is simply the ratio area of triangle 2:area of triangle 4, we then know that this ratio is 2a:5a=2:5.

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.